Page 18 - Elements of Distribution Theory

P1: JZP
052184472Xc01 CUNY148/Severini May 24, 2005 17:52

4 Properties of Probability Distributions

Suppose (P5) holds. Then

P lim A c n = lim P A c n
n→∞ n→∞
so that
c

c
P lim A n = 1 − P lim A n = 1 − lim P A n = lim P(A n ),
n→∞ n→∞ n→∞ n→∞
proving (P4). A similar argument may be used to show that (P4) implies (P5). Hence, it
suffices to show that (P3) and (P4) are equivalent.
Suppose A 1 , A 2 ,... is an increasing sequence of events. For n = 2, 3,..., define
¯
A n = A n ∩ A c n−1 .
Then, for 1 < n < k,
¯
¯

A n ∩ A k = (A n ∩ A k ) ∩ A c n−1 ∩ A c k−1 .
Note that, since the sequence A 1 , A 2 ,... is increasing, and n < k,
A n ∩ A k = A n
and
A c ∩ A c = A c .
n−1 k−1 k−1
Hence, since A n ⊂ A k−1 ,
¯
¯
A n ∩ A k = A n ∩ A c =∅.
k−1
Suppose ω ∈ A k . Then either ω ∈ A k−1 or ω ∈ A c ¯
k−1 ∩ A k = A k ; similarly, if ω ∈ A k−1
¯
c
then either ω ∈ A k−2 or ω ∈ A ∩ A k−1 ∩ A c k−2 = A k−1 . Hence, ω must be an element of
1
¯
¯
¯
either one of A k , A k−1 ,..., A 2 or of A 1 . That is,
¯
¯
¯
A k = A 1 ∪ A 2 ∪ A 3 ∪· · · ∪ A k ;
¯
hence, taking A 1 = A 1 ,
k

¯
A k = A n
n=1
and
∞

¯
lim A k = A n .
k→∞
n=1
Now suppose that (P3) holds. Then

∞ ∞ k
¯
¯
¯

P( lim A k ) = P A n = P(A n ) = lim P(A n ) = lim P(A k ),
k→∞ k→∞ k→∞
n=1 n=1 n=1
proving (P4).
Now suppose that (P4) holds. Let A 1 , A 2 ,... denote an arbitrary sequence of disjoint
subsets of and let
∞

A 0 = A n .
n=1